4
4.1
Schrödinger Equation
Dynamics of Schrödinger’s ψ-function
Schrödinger wrote down guess for wave equation for a particle in external
potential based on de Broglie’s wave idea. Noted that free particle was
represented by lin. comb. of plane waves
ei(k·r−ωt),
h̄ω = h̄2k 2/2m
(1)
Any lin. comb. ψ of this type satisfies diff. eqn. with linear time
deriviative:
∂ψ −h̄2 2
=
∇ψ
(2)
ih̄
∂t
2m
de Broglie said the particle’s energy E is h̄ω, therefore
ih̄
∂ψ
= h̄ωψ = Eψ =⇒
∂t
−h̄2 2
∇ ψ = Eψ
2m
(3)
(4)
LHS of (4) looks a little like the Hamiltonian of the free particle, p2/2m
in classical physics, if we were to write
p ≡ −i∇ =⇒ Hψ = Eψ
(5)
Now what if particle moves in potential V (r)? S. guessed
−h̄2 2
Hψ = (
∇ ψ + V (r)) = Eψ
2m
1
(6)
and if the particle is not in a state of definite energy E S. guessed the
most general equation would in fact be
? ? ? ih̄ ∂ψ
∂t = Hψ,
H=
−h̄2 2
2m ∇
+ V (r)
? ??
(7)
This eqn. is the basis for much of this course. Importance: although a
guess, it systematized quantum theory, which until then was based on a
grabbag of unrelated guesses, or quantization recipes for different situations.
4.2
Hydrogen atom
Take proton to be fixed classical pt. charge, so potential e− feels is
V (r) = −e2/4π²0r,
(8)
and wave eqn. for electron wave fctn. is
−h̄2 2
∇ ψ + V (r)ψ = Eψ,
(9)
2m
assuming the electron to be in a definite energy state w/ energy E. Assume
√ 2
soln. dependent only on r = x + y 2 + z 2, spherically symmetric. We’ll
need following quantities:
∂
∂
∂
+ ĵ
+ k̂ ,
∂x
∂y
∂z
∂2
∂2
∂2
2
∇ = 2 + 2 + 2,
∂x
∂y
∂z
∂ψ dψ ∂r x dψ
=
=
,
∂x
dr ∂x r dr
∂ 2ψ x2 d2ψ 1 dψ x2 dψ
= 2 2 +
− 3 ,
∂x2
r dr
r dr
r dr
∇ = î
2
(10)
(11)
(12)
(13)
d2ψ 2 dψ
∇ψ= 2 +
dr
r dr
2
so
(14)
and S’s eqn. becomes
d2ψ 2 dψ 2m
e2
+
= 2 (E −
)ψ
dr2 r dr
4π²0r
h̄
(15)
where E = −E is binding energy (expect E < 0!).
Now use std. trick to solve eqns. of this type. Put ψ = f (r)e−αr , find
2 0 2α
2mE
2e2m f
f − 2αf + α f + f − f = 2 f −
r
r
h̄
4π²0h̄2 r
00
0
2
(16)
Choose α2 = 2mE/h̄2 to get rid of 2 underlined terms→
2 0 2α
2e2m f
f − 2αf + f − f = −
r
r
4π²0h̄2 r
00
0
(17)
Now expand f in power series
f=
∞
X
p=0
Ap[p(p−1)r
p−2
−2αpr
p−1
X
p
+2pr
Aprp,
p−2
−2αr
(18)
p−1
2me2 p−1
r ] = 0 (19)
+
4π²0h̄2
Since r is arbitrary, coeff. of rp−1 has to vanish! This gives recursion
relation
2me2
Ap+1((p + 1)p + 2(p + 1)) = Ap(2α(p + 1) −
4π²0h̄2
3
(20)
or
Ap+1 2α(p + 1) − 2me2/(4π²0h̄2)
=
Ap
p2 + 3p + 2
(21)
Suppose the sequence doesn’t terminate, i.e. numerator doesn’t vanish for
any integer p ≥ 0, then as p → ∞ this becomes
Ap+1
2α
(2α)p
=⇒ Ap ∼
∼
Ap
p
p!
(22)
f ∼ e2αr =⇒ ψ ≡ e−αr f ∼ eαr ,
(23)
or
major bad news since ψ would diverge at ∞.
So instead suppose numerator vanishes for some p ≡ n − 1:
me2 1
α =
4π²0h̄2 n
v
u
u 2mE
u
= t 2
h̄
(24)
(25)
Or, lo and behold,
En =
m
2h̄2 n2
µ
e2
4π²0
¶2
(26)
agrees with Bohr’s result and spectroscopy. Great success! Recall we
assumed ψ = ψ(r) only. Turns out there are anisotropic solns as well,
ψ(r, θ, φ), but they don’t add any new En. Will revisit.
4.3
1D Simple Harmonic Oscillator
Classically, H = p2/2m + kx2/2. So according to S.’s prescription, in
quantum version should write
h̄2 ∂ 2ψ 1 2
Hψ = −
+ kx ψ = Eψ
(27)
2m ∂x2 2
4
From here on out as usual there is nothing mysterious, only cookbook
techniques to solve differential equations. We’ll use ladder operator technique, because it allows us to discuss quantum states in somewhat intuitive
way. Define operators
v
u
u
u
t
k
h̄ ∂
x− √
2
2m ∂x
v
u
uk
h̄ ∂
u
L− ≡ t x + √
2
2m ∂x
L+ ≡
(28)
(29)
and note that product is
2
2
v
u
u
u
t
h̄ ∂
h̄ k
∂
∂
k
+
(x
−
x)
L+ L− = x2 −
2
2m ∂x2 2 m ∂x ∂x
(30)
∂
WARNING: recall this is a differential operator. So when I write ∂x
x
it does not mean 1, but rather is defined such that when applied to a
function ψ(x) you get
∂
∂
∂ψ
xψ=
(xψ) = ψ + x
∂x
∂x
∂x
Using this rule, easy to see that for any fctn. ψ
5
(31)
∂
∂
xψ − x ψ = ψ,
∂x
∂x
(32)
or,
∂
∂
x−x )=1
∂x
∂x
which we sometimes write in even more compact notation
(
[
∂
, x] = 1
∂x
(33)
(34)
[A, B] = AB − BA is called the commutator of two differential operators
(or noncommuting matrices!)
∂
Now using [ ∂x
, x] = 1 in Eq. (30), the product becomes
2
L+L− =
2
v
u
u
u
t
k 2 h̄ ∂
h̄ k
x −
−
2m ∂x2} |2 {zm}
|2
{z
h̄
ω
H
2
(35)
where ω is classical oscillation freq.
L+L− = H − h̄ω/2
(36)
Similar calc. gives (check!)
L−L+ = H + h̄ω/2
(37)
[L+, L−] = −h̄ω
(38)
so we have shown
Now compute commutator of L+ and H:
6
[L+, H] = [L+, L+L− + h̄ω/2]
= [L+, L+L−] + [L
+ , h̄ω/2]
|
{z
}
=0
= L+L+L− − L+L−L+
= L+[L+, L−]
= −h̄ωL+
(39)
and similarly
[L−, H] = h̄ωL−
(40)
We can now use this algebra of L’s to find the allowed energies E. Suppose
E is a solution (eigenvalue) corresponding to a solution ψ to S.’s eqn.
Operate on both sides of eqn. by L+:
L+Hψ = L+Eψ
([L
+{z, H]} +HL+ )ψ = EL+ ψ
|
(41)
− h̄ωL+
or
H(L+ψ) = (E + h̄ω)(L+ψ)
(42)
So L+ψ is an eigenfunction (i.e. solution) with eigenvalue E + h̄ω. L+
is called a ladder operator or raising operator, because it “lifts” ψ to a
new ψ 0 one quantum of energy h̄ω up the “ladder” of energies. Check that
L− is lowering operator, i.e. it gives new wavefunction with eigenvalue
reduced by h̄ω.
? ? ? Note E can’t be lowered indefinitely, at least we suspect so since
in classical physics energy is positive definite, p2/2m + kx2/2 ≥ 0. For
the moment, assume E ≥ 0 and ask how the sequence of lowerings might
be terminated. Ladder of lowerings stops if we ever get
7
L−ψ0 = 0
(43)
for some ψ0. But we figure out the energy corresponding to ψ0 by expressing H as in Eq. (36):
Hψ0 = (L
+ L− +h̄ω/2)ψ0
| {z }
0 since L−ψ0 = 0
h̄ω
=
ψ0.
2
(44)
So the putative lowest energy state has energy E0 ≡ h̄ω/2. Now apply
raising operator n times:
H(L+)nψ0 = (E0 + nh̄ω)(L+)nψ0
(45)
So we get just the spacing of levels Planck assumed, with the one minor
change that there is a it zero-point energy) h̄ω/2 by which they are all
offset:
En = h̄ω(n + 1/2)
(46)
Q: why doesn’t zero-pt. energy affect specific heat calculation?
A: depends only on the density of levels, at least in a large system!
4.3.1
1D SHO eigenfunctions
Now let’s look at the eigenfunctions of the SHO. We’ve constructed a
ladder, so let’s start at the bottom. The argument was that if I take, e.g.
the lowering operator L− and apply it to an eigenstate ψn (eigenvalue En),
I will get another eigenstate corresponding to En−1. Well this can’t go on
forever. Physically we expect that the spectrum of En must be bounded
below, by zero certainly, since the classical Hamiltonian is positive definite!
So the only possibility to cut off the sequence of lowerings is if for some
8
eigenfunction ψ we find that L−ψ = 0. This must be the lowest energy
eigenfunction, the so called ground state eigenfunction of the SHO, so
we’ll start indexing from here and call the wave function ψ0. To find its
form we must solve the differential equation
v
u
u
u
t


k
h̄ ∂ 
 ψ (x) = 0.
x+ √
 0
2
2m ∂x
(47)
Now let’s guess the solution. You might look for a wave function which
is centered around the origin x = 0, and which falls off rapidly very far
from 0 because there is very little chance of finding the particle there if
it has very low energy (classically, very small amplitude!). You might
guess ψ0 ∼ e−α|x|, for example, an exponential decay in both positive
and negative directions. However if you subsitute (e.g. positive x) you
find the first term gives you xe−αx and the second one e−αx, except for
some constant factors. So the two terms can’t cancel for all x, & this just
2
isn’t a solution. How about a Gaussian, e−αx ? √
This works, as you can
verify by direct substitution, if you choose α = mk/(2h̄). Check the
√
dimensions– α is an inverse length, so this rdefines a characteristic length
in the problem, α−1/2 = (h̄2/(mk))1/4 = h̄/(mω), which is a kind of
quantum mechancical “amplitude” of oscillation in the ground state, as
we will see in the problem set.
2
Note any function Ae−αx satisfies the lowering condition. We will want
to normalize the wave functions such that
Z
dx|ψ|2 = 1,
(48)
so plugging this in and using
Z ∞
e
−∞
−2αy 2
=
v
u
u
t
π
,
2α
(49)
which gives A = (mω/(h̄π))1/4.
Finally, we know that we can get all the other eigenfunctions in the
“ladder” now by applying L+ one function at a time, starting with ψ1 ∝
9
L+ψ0, etc. Note that I used ∝ instead of =, since in each case we should
apply the normalization condition. In general the solution to the SHO
differential equation is proportional to
2
ψn(x) = Hn(x)e−αx ,
where Hn is a Hermite polynomial. See Griffiths p. 56.
10
(50)